PLEASE HELPPP !!! The half-life of a specific radioactive compound is 20 years. The equation used to model this half-life is y = 5(1/2) t/20, where t represents time. Currently, there is 5 pounds of this compound below the earth's surface. When will there only be 1 pounds of this compound left? Round your answer to the nearest year.

Question

PLEASE HELPPP !!! The half-life of a specific radioactive compound is 20 years. The equation used to model this half-life is  y = 5(1/2) t/20, where t represents time. Currently, there is 5 pounds of this compound below the earth's surface. When will there only be 1 pounds of this compound left? Round your answer to the nearest year.

campbell_st · Answer

so is the equation 

$$y = 5(\frac{1}{2})^{\frac{t}{20}}$$

anonymous · Answer

@ziko1995 thats wrong

anonymous · Answer

and yes it is

campbell_st · Answer

I think you need to find the value of t when 

$$1 = 5(\frac{1}{2})^{\frac{t}{20}}$$

so its the same method as above by use ln(0.2) instead of ln(0.8)

campbell_st · Answer

so you have 

$$0.2 = (\frac{1}{2})^{\frac{t}{20}}$$

anonymous · Answer

0.2??

campbell_st · Answer

because you have 1 pound left

so 

1 = 5(1/2)^{t/20}

divide both sides of the equation by 5

0.2 = (1/2)^{t/20}

take the log of both sides

ln(0.2) = t/20ln(1/2)

divide both sides by ln(1/2)

ln(0.2)/ln(1/2) = t/20

multiply both sides by 20

20*ln(0.2)/ln(1/2) = t

just evaluate for the amount of time

campbell_st · Answer

so half line means it taks 20 years to go from 5lb to 2.5lb.

then another 20 years to go from 2.5 lb to 1.25 lb... so the answer makes sense.

anonymous · Answer

its wrong

anonymous · Answer

the answer can't be 20, if it's in the question

campbell_st · Answer

o well so 46 ot 47 years is wrong...

campbell_st · Answer

do you know the correct answer?

anonymous · Answer

no it does not sa it says this when i get it wrong Put the given equation in y1 of your calculator.  Put the amount remaing in y2.  Find the intersection to determine the number of years until the amount in y2 is remaining.

campbell_st · Answer

well so you should enter 

$$y1 = 5\times(1/2)^{(x/20)}$$

and 
y2 = 1 

then reset the window so you can see the graph, 

press 2nd CALC and select intersection. 

but I haven't used a graphics calc for ages

but here is a graph. I've rewritten the equation in x and y rather than y and t

campbell_st · Answer

and it shows t = 46.3.... years

anonymous · Answer

In the question is says to round so should i try that answer

anonymous · Answer

yayyy its right thanks so much :)

anonymous · Answer

I followed ur steps for this problem that's similar but can't figure out the answer!! Can you help ??? The half-life of a specific radioactive compound is 20 years. The equation used to model this half-life is  y = 5(1/2)t/20 , where t represents time. Currently, there are 5 pounds of this compound 15 feet below the earth's surface. When will there only be 4 pounds of this compound left? Round your answer to the nearest tenth of a year.